Question

A train is slowing down with an average acceleration of -6.0 m/s^2. If its initial velocity is 60.0 m/s, how far does it travel in 5 seconds? A) 150 m B) 180 m C) 210 m D) 240 m

Answer 1

The train travels a distance of 292.5 meters in 5 seconds.

Step-by-step explanation:

The distance covered by the train can be calculated using the equation:

d = vi*t + (1/2)*a*t²

Where d is the distance, vi is the initial velocity, t is the time, and a is the acceleration.

In this case, the initial velocity, vi, is 60.0 m/s and the acceleration, a, is -6.0 m/s² (negative because the train is slowing down).

Substituting the values into the equation, we have:

d = 60.0*5 + (1/2)*(-6.0)*(5²)

Simplifying the equation, we get:

d = 300 + (-7.5)

d = 292.5 m

Therefore, the train travels a distance of 292.5 meters in 5 seconds.

Answer 2

The closest answer to this calculation is : C) 210 m.

To calculate the distance traveled by the train, we can use the formula for uniformly accelerated motion:

`\[ d = v_i t + (1)/(2) a t^2 \]`

where:

-
`\( d \)` is the distance traveled,

-
`\( v_i \)` is the initial velocity,

-
`\( a \)` is the acceleration,

-
`\( t \)` is the time.

Given:

-
`\( v_i = 60.0 \, \text{m/s} \),`

-
`\( a = -6.0 \, \text{m/s}^2 \)` (negative because the train is decelerating),

-
`\( t = 5 \, \text{s} \).`

Plug these values into the formula:

`\[ d = 60.0 * 5 + (1)/(2) * (-6.0) * 5^2 \]`

`\[ d = 300 + (1)/(2) * (-6.0) * 25 \]`

`\[ d = 300 - 3 * 25 \]`

`\[ d = 300 - 75 \]`

`\[ d = 225 \, \text{m} \]`